3.1054 \(\int \frac {\cosh ^2(a+b x)-\sinh ^2(a+b x)}{\cosh ^2(a+b x)+\sinh ^2(a+b x)} \, dx\)

Optimal. Leaf size=11 \[ \frac {\tan ^{-1}(\tanh (a+b x))}{b} \]

[Out]

arctan(tanh(b*x+a))/b

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Rubi [A]  time = 0.06, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {4380, 203} \[ \frac {\tan ^{-1}(\tanh (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[(Cosh[a + b*x]^2 - Sinh[a + b*x]^2)/(Cosh[a + b*x]^2 + Sinh[a + b*x]^2),x]

[Out]

ArcTan[Tanh[a + b*x]]/b

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 4380

Int[(u_.)*((a_.) + cos[(d_.) + (e_.)*(x_)]^2*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)]^2)^(p_.), x_Symbol] :> Dist
[(a + c)^p, Int[ActivateTrig[u], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[b - c, 0]

Rubi steps

\begin {align*} \int \frac {\cosh ^2(a+b x)-\sinh ^2(a+b x)}{\cosh ^2(a+b x)+\sinh ^2(a+b x)} \, dx &=\int \frac {1}{\cosh ^2(a+b x)+\sinh ^2(a+b x)} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac {\tan ^{-1}(\tanh (a+b x))}{b}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 1.55 \[ \frac {\tan ^{-1}(\sinh (2 a+2 b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[a + b*x]^2 - Sinh[a + b*x]^2)/(Cosh[a + b*x]^2 + Sinh[a + b*x]^2),x]

[Out]

ArcTan[Sinh[2*a + 2*b*x]]/(2*b)

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fricas [B]  time = 0.40, size = 38, normalized size = 3.45 \[ -\frac {\arctan \left (-\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(b*x+a)^2-sinh(b*x+a)^2)/(cosh(b*x+a)^2+sinh(b*x+a)^2),x, algorithm="fricas")

[Out]

-arctan(-(cosh(b*x + a) + sinh(b*x + a))/(cosh(b*x + a) - sinh(b*x + a)))/b

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giac [B]  time = 0.16, size = 44, normalized size = 4.00 \[ -\frac {\arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (b x + a\right )}\right )}\right ) - \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (b x + a\right )}\right )}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(b*x+a)^2-sinh(b*x+a)^2)/(cosh(b*x+a)^2+sinh(b*x+a)^2),x, algorithm="giac")

[Out]

-(arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^(b*x + a))) - arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^(b*x + a))))/b

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maple [B]  time = 0.63, size = 148, normalized size = 13.45 \[ \frac {2 \sqrt {2}\, \arctan \left (\frac {2 \tanh \left (\frac {b x}{2}+\frac {a}{2}\right )}{-2+2 \sqrt {2}}\right )}{b \left (-2+2 \sqrt {2}\right )}-\frac {2 \arctan \left (\frac {2 \tanh \left (\frac {b x}{2}+\frac {a}{2}\right )}{-2+2 \sqrt {2}}\right )}{b \left (-2+2 \sqrt {2}\right )}-\frac {2 \sqrt {2}\, \arctan \left (\frac {2 \tanh \left (\frac {b x}{2}+\frac {a}{2}\right )}{2+2 \sqrt {2}}\right )}{b \left (2+2 \sqrt {2}\right )}-\frac {2 \arctan \left (\frac {2 \tanh \left (\frac {b x}{2}+\frac {a}{2}\right )}{2+2 \sqrt {2}}\right )}{b \left (2+2 \sqrt {2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(b*x+a)^2-sinh(b*x+a)^2)/(cosh(b*x+a)^2+sinh(b*x+a)^2),x)

[Out]

2/b*2^(1/2)/(-2+2*2^(1/2))*arctan(2*tanh(1/2*b*x+1/2*a)/(-2+2*2^(1/2)))-2/b/(-2+2*2^(1/2))*arctan(2*tanh(1/2*b
*x+1/2*a)/(-2+2*2^(1/2)))-2/b*2^(1/2)/(2+2*2^(1/2))*arctan(2*tanh(1/2*b*x+1/2*a)/(2+2*2^(1/2)))-2/b/(2+2*2^(1/
2))*arctan(2*tanh(1/2*b*x+1/2*a)/(2+2*2^(1/2)))

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maxima [B]  time = 0.45, size = 49, normalized size = 4.45 \[ \frac {\arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (-b x - a\right )}\right )}\right ) - \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (-b x - a\right )}\right )}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(b*x+a)^2-sinh(b*x+a)^2)/(cosh(b*x+a)^2+sinh(b*x+a)^2),x, algorithm="maxima")

[Out]

(arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^(-b*x - a))) - arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^(-b*x - a))))/b

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mupad [B]  time = 0.09, size = 25, normalized size = 2.27 \[ \frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)^2 - sinh(a + b*x)^2)/(cosh(a + b*x)^2 + sinh(a + b*x)^2),x)

[Out]

atan((exp(2*a)*exp(2*b*x)*(b^2)^(1/2))/b)/(b^2)^(1/2)

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sympy [A]  time = 2.24, size = 56, normalized size = 5.09 \[ \begin {cases} \frac {x \left (- \sinh ^{2}{\relax (a )} + \cosh ^{2}{\relax (a )}\right )}{\sinh ^{2}{\relax (a )} + \cosh ^{2}{\relax (a )}} & \text {for}\: b = 0 \\- x & \text {for}\: a = \log {\left (- i e^{- b x} \right )} \vee a = \log {\left (i e^{- b x} \right )} \\\frac {\operatorname {atan}{\left (\frac {\sinh {\left (a + b x \right )}}{\cosh {\left (a + b x \right )}} \right )}}{b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((cosh(b*x+a)**2-sinh(b*x+a)**2)/(cosh(b*x+a)**2+sinh(b*x+a)**2),x)

[Out]

Piecewise((x*(-sinh(a)**2 + cosh(a)**2)/(sinh(a)**2 + cosh(a)**2), Eq(b, 0)), (-x, Eq(a, log(I*exp(-b*x))) | E
q(a, log(-I*exp(-b*x)))), (atan(sinh(a + b*x)/cosh(a + b*x))/b, True))

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